Two infinitely long straight wires lie in the xy - plane along the lines x=±R. The wire located at x = +R carries a constant current I1 and the wire located at x =-R carries a constant current I2. A circular loop of radius R is suspended with its centre at 0, 03R and in a plane parallel to the xy - plane. This loop carries a constant current I in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the +j^ direction. Which of the following statements regarding the magnetic field B→ is (are) true?

(A) If I1=I2, then B→ cannot be equal to zero at the origin 0, 0, 0

(B) If I1>0 and I2<0, then B→ can be equal to zero at the origin 0, 0, 0

(C) If I1 0, then B→ can be equal to zero at the origin 0, 0, 0

(D) If I1=I2 , Then the z-component of the magnetic field at the centre of the loop is (-\mu 0 \mathrm{I} / 2 \mathrm{R})

JEE Advanced · Physics · 20. Magnetism & Current

Two infinitely long straight wires lie in the xy - plane along the lines $x = \pm R$ . The wire located at $x = + R$ carries a constant current $I_{1}$ and the wire located at $x = - R$ carries a constant current $I_{2}$ . A circular loop of radius R is suspended with its centre at $(0, 0 \sqrt{3} R)$ and in a plane parallel to the xy - plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+ \hat{j}$ direction. Which of the following statements regarding the magnetic field $\vec{B}$ is (are) true?

A If $I_{1} = I_{2}, t h e n \vec{B}$ cannot be equal to zero at the origin $(0, 0, 0)$
B If $I_{1} > 0 a n d I_{2} < 0, t h e n \vec{B}$ can be equal to zero at the origin $(0, 0, 0)$
C If $I_{1} < 0 a n d I_{2} > 0, t h e n \vec{B}$ can be equal to zero at the origin $(0, 0, 0)$
D If $I_{1} = I_{2}$ , Then the z-component of the magnetic field at the centre of the loop is $(-\mu 0 \mathrm{I} / 2 \mathrm{R})$

Verified Solution

Answer & Solution

Correct Answer

(A) If $I_{1} = I_{2}, t h e n \vec{B}$ cannot be equal to zero at the origin $(0, 0, 0)$

Step-by-step Solution

Detailed explanation

(A)

At origin,

\vec{B} = 0

due to two wires if

I_{1} = I_{2}, h e n c e ({\vec{B}}_{n e t})

at origin is equal to

\vec{B}

due to ring, which is non-zero.

(B)

I_{1} > 0 a n d I_{2} < 0, \vec{B}

at origin due to wires will be along

+ \hat{k}

direction and

\vec{B}

due to ring is along

- \hat{k}

direction and hence

\vec{B}

can be zero at origin

(C)

I_{1} < 0 a n d I_{2} > 0, \vec{B}

at origin due to wires is along

- \hat{k}

and also along

- \hat{k}

due to ring, hence

\vec{B}

cannot be zero

(D)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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